Evaluate the iterated integral. $ \int_1^4 \left( \int_1^2 \dfrac{x^2 + 2y}{x} \, dy \right) dx =$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{15}{2} + 3\ln(4)$ (Choice B) B $\dfrac{7}{2} + \ln(4)$ (Choice C) C $2$ (Choice D) D $1 + 2 \ln(2)$
Solution: Evaluate the inner integral: $\begin{aligned} & \int_1^4 \left( \int_1^2 \dfrac{x^2 + 2y}{x} \, dy \right) dx \\ \\ &= \int_1^4 \left[ xy + \dfrac{y^2}{x} \right]_1^2 dx \\ \\ &= \int_1^4 \left( 2x + \dfrac{4}{x} \right) - \left( x + \dfrac{1}{x} \right) dx \\ \\ &= \int_1^4 x + \dfrac{3}{x} \, dx \end{aligned}$ Evaluate the outer integral: $\begin{aligned} \int_1^4 x + \dfrac{3}{x} \, dx &= \dfrac{x^2}{2} + 3 \ln(x) \bigg|_1^4 \\ \\ &= \left( \dfrac{16}{2} + 3\ln(4) \right) - \left( \dfrac{1}{2} + 3 \ln(1) \right) \\ \\ &= \dfrac{15}{2} + 3\ln(4) \end{aligned}$ The answer: $ \int_1^4 \left( \int_1^2 \dfrac{x^2 + 2y}{x} \, dy \right) dx = \dfrac{15}{2} + 3\ln(4)$